The parity of the pion from \bgroup\color{black}$\pi d\to n n$\egroup.


We can determine the internal parity of the pion by studying pion capture by a deuteron, \bgroup\color{black}${\pi} +{d}\rightarrow n+n. $\egroup The pion is known to have spin 0, the deuteron spin 1, and the neutron spin \bgroup\color{black}${1\over 2}$\egroup. The internal parity of the deuteron is +1. The pion is captured by the deuteron from a 1S states, implying \bgroup\color{black}$\ell=0$\egroup in the initial state. So the total angular momentum quantum number of the initial state is \bgroup\color{black}$j=1$\egroup.

So the parity of the initial state is

\begin{displaymath}\bgroup\color{black}(-1)^\ell P_\pi P_d = (-1)^0 P_\pi P_d = P_\pi\egroup\end{displaymath}

The parity of the final state is

\begin{displaymath}\bgroup\color{black}P_nP_n (-1)^\ell=(-1)^\ell \egroup\end{displaymath}



Because the neutrons are identical fermions, the allowed states of two neutrons are \bgroup\color{black}$^1S_0$\egroup, \bgroup\color{black}$^3P_{0,1,2}$\egroup, \bgroup\color{black}$^1D_2$\egroup, \bgroup\color{black}$^3F_{2,3,4}...$\egroup The only state with \bgroup\color{black}$j=1$\egroup is the \bgroup\color{black}$^3P_{1}$\egroup state, so \bgroup\color{black}$\ell=1$\egroup

\begin{displaymath}\bgroup\color{black}\Rightarrow P_\pi =-1.\egroup\end{displaymath}

Jim Branson 2013-04-22